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# Missing Number

Updated: Mar 24, 2021

Given an array nums containing n distinct numbers in the range [0, n], return the only number in the range that is missing from the array.

Follow up: Could you implement a solution using only O(1) extra space complexity and O(n) runtime complexity?

Example 1:

```Input: nums = [3,0,1]
Output: 2
Explanation: n = 3 since there are 3 numbers, so all numbers are in the range [0,3]. 2 is the missing number in the range since it does not appear in nums.

```

Example 2:

```Input: nums = [0,1]
Output: 2
Explanation: n = 2 since there are 2 numbers, so all numbers are in the range [0,2]. 2 is the missing number in the range since it does not appear in nums.

```

Example 3:

```Input: nums = [9,6,4,2,3,5,7,0,1]
Output: 8
Explanation: n = 9 since there are 9 numbers, so all numbers are in the range [0,9]. 8 is the missing number in the range since it does not appear in nums.

```

Example 4:

```Input: nums = 
Output: 1
Explanation: n = 1 since there is 1 number, so all numbers are in the range [0,1]. 1 is the missing number in the range since it does not appear in nums.

```

Constraints:

• n == nums.length

• 1 <= n <= 104

• 0 <= nums[i] <= n

• All the numbers of nums are unique.

Solution:

```class Solution {
public int missingNumber(int[] nums) {
int sum1=0,sum2=0;
sum1=(nums.length*(nums.length+1))/2;
for(int i:nums) sum2+=i;
return sum1-sum2;
}
}```